Question

Three workers at a fast food restaurant pack the take-out chicken dinners. john packs 45% of the dinners, mary packs 25% of the dinners and sue packs the remaining dinners. of the dinners john packs 4% do not include a salt packet. if mary packs the dinner 2% of the time the salt is omitted. lastly, 3% of the dinner's do not include salt if sue does the packing. what is the probability that you will have salt packed with your dinner

Answer 1

To calculate this probability is often easier to calculate the complement
the prob that we are looking for is 1-P(No salt packed in dinner)
Note: No salt packed with dinner is the complement event of having salt packed within
so let's find this probability
He is what we know
if John packs 4/45 chance no salt included
if Sue packs 3/70 chance no salt
if Marry packs 2/25 No salt
We are looking into probability is the sum (note these events are disjoint)
final result should be 1- that sum

Answer 2

The probability of that you will have salt packed with your dinner is $\boxed{0.968}.$

Further explanation:

The probability can be obtained as the ratio of favorable number of outcomes to the total number of outcomes.

$\boxed{{\text{Probability}} = \frac{{{\text{Favorable number of outcome}}}}{{{\text{Total number of outcomes}}}}}$

Given:

John packs $45\%$ of the dinners, Mary packs $25\%$ of the dinners and Sue packs the remaining dinners. Of the dinners John packs $4\%$ do not include a salt packet. If Mary packs the dinner $2\%$ of the time the salt is omitted. Lastly, $3\%$ of the dinners do not include salt if Sue does the packing.

Explanation:

The probability of John packs the dinner can be expressed as follows,

$\begin{aligned}{\text{Probability}}\left( {{\text{John}}} \right) &= 45\%\\&= 0.45\\\end{gathered}$

The probability of Mary packs the dinner can be expressed as follows,

$\begin{aligned}{\text{Probability}}\left( {{\text{Mary}}} \right) &= 25\%\\&= 0.25\\\end{aligned}$

The probability of Sue packs the dinner can be expressed as follows,

$\begin{aligned}{\text{Probability}}\left( {{\text{Sue}}} \right) &= 1 - 0.45 - 25\\&= 1 - 0.70\\&= 0.30\\\end{aligned}$

The probability of no salt packet is packed given that the John packs for dinner can be expressed as follows,

$P\left( {\dfrac{{{\text{No salt}}}}{{{\text{John packs}}}}} \right) = 0.04$

The probability of no salt packet is packed given that the Mary packs for dinner can be expressed as follows,

$P\left( {\dfrac{{{\text{No salt}}}}{{{\text{Mary packs}}}}} \right) = 0.02$

The probability of no salt packet is packed given that the Sue packs for dinner can be expressed as follows,

$P\left( {\dfrac{{{\text{No salt}}}}{{{\text{Sue packs}}}}} \right) = 0.03$

The probability of that you will have salt packed with your dinner can be expressed as follows,

$\begin{aligned}{\text{Probability}}\left( {{\text{salt}}} \right) &= 1 - {\text{Probability}}\left( {{\text{No salt}}} \right)\\&= 1 - \left( {0.45 \times 0.04 + 0.25 \times 0.02 + 0.30 \times 0.03} \right)\\&= 1 - 0.032\\&= 0.968\\\end{aligned}$

The probability of that you will have salt packed with your dinner is $\boxed{0.968}.$

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497.

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Probability

Keywords: Three worker, fast food, restaurant pack, chicken dinners, John packs, 45% of the dinners, Mary, 25% of dinner, Sue packs remaining, 4% so not include salt, salt packet, salt is omitted, dinner, 2%, 3%, purchased dinner, probability.